Question

3) Determine power rating of an electric motor if it
runs at 1440 r.p.m and line shaft transmits torque a
75 Nm. Assume Reduction ratio = 1.6​

Answer 1

Answer:

The tendency of force to rotate an object about an axis is called as torque or moment.

Given: speed (n) : 1440 r.p.m, torque = 75 Nm, reduction ratio (I) = 1.6

Formula: P = 2 p N T / 60

Solution:

i = n / N

1.6 = 1440 / N

N = 900 r.p.m

P = (2 p N T)/ 60

P = (2 x 3.14 x 900 x 75 ) /60

P = 423900 /60

= 7065 W

= 7.065 kW

Power rating of an electric motor is 7.065 kW

Answer 2

Answer:

Power rating of an electric motor is 7.065 kW.

Explanation:

Given, Speed = 1440 r.p.m,Torque =75 Nm,Reduction ratio is 1.6

We know,

$P = \frac{2 \times \pi \times N \times T}{60}$

Reduction ratio = $\frac{n}{N}$

→1.6$= \frac{1440}{N}$

N$= 900 r.p.m$

$P = \frac{2 \times \pi \times NT}{60} = \frac{2 \times 3.14 \times 900 \times 75}{60} = \frac{423900}{60} = 7065 W = 7.065 kW$

Power rating of an electric motor is 7.065 kW.